when a subgroup is in the center of other subgroup

The question is

Let G be a group, $\displaystyle N \lhd G $ such that $\displaystyle \[G:N\] = r $

let $\displaystyle H \lhd G $ with finite order show that

if

$\displaystyle H \cap N = e $ then $\displaystyle nh = hn \;\; \forall \; h\in H \; , n\in N $

Re: when a subgroup is in the center of other subgroup

i am a little unsure as to how [G:N] = r is relevant to your question, other than saying N is of finite index. i also don't know what you mean by "one subgroup is in the center of the other" in terms of your subsequent question.

if H is in Z(N), of course nh = hn, but then it cannot be the case that H∩N = {e}, unless Z(N) is likewise trivial.

but if H,N are both normal in G, with H∩N = {e}, then for any h in H, and n in N, we have:

$\displaystyle hnh^{-1}n^{-1} \in N$ since $\displaystyle hnh^{-1}, n^{-1} \in N$, because N is normal, while

$\displaystyle hnh^{-1}n^{-1} \in H$, since $\displaystyle h, nh^{-1}n^{-1} \in H$ because H is normal.

since H∩N = {e}, we see that h and n commute.

Re: when a subgroup is in the center of other subgroup

thanks

the problem in Abstract algebra when you see the answer you say oooh it is easy how i did not figure that

the hardness is the idea of the solution

I think if I had thought in it in reverse nh = hn and see what will happen

anyway thanks very much